Height and Distance solved questions - WTSkills- Learn Maths, Quantitative Aptitude, Logical Reasoning

This is another post from the quantitative aptitude chapter Height and Distance.
All the questions are solved step by step with the illustrated diagram. If you want to develop the skill of solving height and distance problems, there are two things you need to practice:

a. Properties of Triangle
b. Trigonometry Functions

As the chapter involves finding height, distance and angle of elevation of any platform, you will need to draw a triangle for each and every question. After that concept of trigonometry is used to get the required data.

I strongly suggest to clear the above two concept first and after that you can try to attempt the below questions

Height and Distance Quantitative Aptitude

Q1. The angle of elevation of the top of a tower from a point on the ground is a 30º and moving 70 meters towards the tower it becomes 60º. The height of the tower is

(a) $10\sqrt { 3 }$ (b) $11\sqrt { 3 }$

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height and distance quantitative aptitude questions for competition exams like GMAT, GRE, Math Olympiad, CAT, CMAT, NMAT, SNAP, SSC, SSC-CGL, SSC-CHSL, Banking exams, sbi po, sbi clerk, NDA, AFCAT

Solution
CD = h
BC = x

In\quad \triangle BCD\\ \\ tan60=\frac { CD }{ BC } \\\ \\ \sqrt { 3 } =\frac { h }{ x } \\\ \\ h=x\sqrt { 3 } \quad \quad \quad ….(a)\\\ \\ In\quad \triangle ACD\\ \\ tan30=\frac { CD }{ AC } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x+20 } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { x\sqrt { 3 } }{ x+20 } \quad \quad \quad \quad \quad [h=x\sqrt { 3 } ]\\\ \\ x=10\\ \\ h=x\sqrt { 3 } =10\sqrt { 3 } \quad m

Option (a) is the right answer

Q2. From an aeroplane just over a river are found to be 60º and 30º respectively. If the breadth of the river is 400 meter, then the height of the aeroplane above the river at that instant is (Assume =1.732)

(a) 519.2 m (b) 346.4 m

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Solution

important height and distance solved questions

BC = 400 meter
AD = h
BD = x

In\quad \triangle ABD\\ \\ tan60=\frac { AD }{ BD } \\\ \\ \sqrt { 3 } =\frac { h }{ x } \\\ \\ x=\frac { h }{ \sqrt { 3 } } \quad ……(a)\\\ \\ \\\ \\ In\quad \triangle ACD\\ \\ tan30=\frac { AD }{ CD } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 400-x } \\\ \\ \sqrt { 3h } =400-x\\\ \\ \sqrt { 3h } =400-\frac { h }{ \sqrt { 3 } } \quad \quad [x=\frac { h }{ \sqrt { 3 } } ]\\\ \\ \ h=\frac { 400\sqrt { 3 } }{ 4 } \\\ \\ h=100\sqrt { 3 } \\\ \\ h=100\times 1.732=173.2\quad meter

option (c) is the right answer

Q3. The angle of elevation of the sun when the length of the shadow of a pole is equal to its height is:

(a) 30º (b) 45º

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Solution:

how to solve height and distance aptitude questions

AB = h
Shadow = BC
AB= BC= h

Angle of elevation = θ
tan θ = AB/BC
tan θ = h/h
tan θ = 1
so θ = 45

Hence angle of elevation is 45 degree

option (b) is the right answer

Q4. A person observes that the angle of elevation at the top of a pole of height 5 meter is 30º. Then the distance of the person from the pole is:

(a) 5 m (b) 10 m

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Solution

collection of height and distance questions for aptitude exams

AB = 5 meter
tan 30 = AB/BC

$\frac { 1 }{ \sqrt { 3 } } =\frac { 5 }{ BC } \\\ \\ BC=5\sqrt { 3 } m\\\ \\ Distance\quad between\quad person\quad and\quad pole\quad is\quad 5\sqrt { 3 } \quad meter.$

option (d) is the right answer

Q5. The cliff of a mountain is 180 m high, and the angles of depression of two ships on the either side of cliff are 30º and 60º. What is the distance between the two ships?

(a) 400 m (b) 400 m

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Solution:

AD = 180 meter
Angle ABD = 60 degree
Angle ACD = 30 degree

In\quad \triangle ABD\\ \\ tan60=\frac { AD }{ BD } \\\ \\ \sqrt { 3 } =\frac { 180 }{ BD } \\\ \\ BD=\frac { 180 }{ \sqrt { 3 } } =\frac { 60\times 3 }{ \sqrt { 3 } } =\frac { 60\times \sqrt { 3 } \times \sqrt { 3 } }{ \sqrt { 3 } } \quad \\\ \\ BD=60\sqrt { 3 } \quad m \\\ \\ \\\ \\ \\\ \\

In\quad \triangle ACD\\ \\ tan30=\frac { AD }{ CD } \\\ \\ \frac { 1 }{ \sqrt { 3 } } =\frac { 180 }{ CD } \\\ \\ CD=180\sqrt { 3 } \\\ \\ BC=BD+CD\\ \\ BC=60\sqrt { 3 } +180\sqrt { 3 } \\ \\ BC=240\sqrt { 3 } \\ \\ BC=240\times 1.732\\ \\ BC=415.68\quad m\\\ \\ Distance\quad between\quad two\quad Ships\quad is\quad 415.68\quad meter.\ \

option (c) is the right answer

Q6. If the angle of elevation of the Sun changes from to , the length of the shadow of a pillar decreases by 20 meters. The height of the pillar is:

(a) $20(\sqrt { 3 } -1)$ (b) $20(\sqrt { 3 } +1)$

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height and distance quantitative aptitude questions for competition exams like GRE, GMAT, Math Olympiad, CAT, MAT, CMAT, NMAT, SNAP, SSC, SSC-CGL, SSC-CHSL, Bank exams, RRB, SBI po, SBI clerk, RRB, NDA, AFCAT

Solution
Let AB be the pillar of height h meter
if BD = x
and DC = 20 meter

Then
BC = BD + DC
BC = (x + 20) meter

From\quad \Delta ABD\\ \\ tan\quad 45\quad =\frac { AB }{ BC } =\frac { h }{ x } \\\ \\ =>\quad 1=\frac { h }{ x } \quad =>\quad h=x\quad …….(a)\\\ \\ \\\ \\ From\quad \Delta ABC,\quad tan30\quad =\frac { AB }{ BC } \\ \\ =>\quad \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x+20 } \\\ \\ =>\quad \sqrt { 3 } h\quad =\quad h+20\quad \quad \quad (h=x)\\\ \\ =>\quad h=\frac { 20 }{ \sqrt { 3 } -1 } \\\ \\ =>\quad h=\frac { 20 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \\\ \\ =>\quad h=\frac { 20(\sqrt { 3 } +1) }{ 3-1 } \quad =\frac { 20(\sqrt { 3 } +1) }{ 2 } \\\ \\ =>\quad h=\quad 10(\sqrt { 3 } +1)\quad meter

option (c) is the right answer

Q7. At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is . On walking 138 meters towards the monument the secant of the angle of elevation is found to be . The height of the monument (in meters) is

(a) 42 (b) 49

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height and distance aptitude questions step by step solution

Solution
AB = height of monument
DC = 138
BD = x

an\quad \alpha =\frac { 1 }{ 5 } \\ \\ sec\quad \beta =\frac { \sqrt { 193 } }{ 12 } \\\ \\ tan\quad \beta =\sqrt { { sec }^{ 2 }\beta -1 } =\sqrt { \frac { 193 }{ 144 } -1 } =\sqrt { \frac { 193-144 }{ 144 } } \\\ \\ tan\quad \beta =\sqrt { \frac { 49 }{ 144 } } =\frac { 7 }{ 12 } \\\ \\ \\\ \\ From\Delta ABC,\quad \ tan\quad \alpha =\frac { AB }{ BC } \\ \\ =>\quad \frac { 1 }{ 5 } =\frac { h }{ x+138 } \\\ \\ =>\quad h=\frac { x+138 }{ 5 } \quad \\\ \\ =>\quad 5h\quad =\quad x+138\quad ….(a)\\\ \\ \\\ \\ From\quad \Delta ABD,\\ \\ tan\quad \beta =\quad \frac { h }{ x } \\\ \\ =>\quad \frac { 7 }{ 12 } =\frac { h }{ x } \\\ \\ =>\quad x=\frac { 12h }{ 7 } \quad \quad ….(b)\\\ \\ \\\ \\ Put\quad value\quad of\quad x\quad in\quad equation\quad (a)\\ \\ 5h=\frac { 12h }{ 7 } +138\\\ \\ 5h=\frac { 12h+(138\times 7) }{ 7 } \quad \\\ \\ 35h-12h\quad =\quad 138\times 7\\\ \\ 23h=138\times 7\\\ \\ h=\frac { 138\times 7 }{ 23 } =42\quad meter.

option (a) is the right answer

Q8. The distance between two pillars of length 16 meters and 9 meters is x meters. If two angles of elevation of their respective top from the bottom of the of the other are complementary to each other then the value of x (in meters) is

(a) 15 (b) 16

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how to solve height and distance problems

Solution
$From\quad \Delta ABC\quad \\ \\ tan(90-\theta )=\frac { AB }{ BC } =\frac { 16 }{ x } \\ \\ =>\quad cot\theta =\frac { 16 }{ x } \quad …..(a)\\\ \\ \\\ \\ From\quad \Delta BCD\\ \\ tan\theta =\frac { DC }{ BC } =\frac { 9 }{ x } ……(b)\quad \\\ \\ \\\ \\ tan\theta \times cot\theta =\frac { 9 }{ x } \times \frac { 16 }{ x } \\ \\ 1=\frac { 16\times 9 }{ { x }^{ 2 } } \quad \quad \quad \quad [tan\theta \times cot\theta =1]\\\ \\ { x }^{ 2 }=16\times 9\\\ \\ x=\sqrt { 16\times 9 } =4\times 3=12\quad meter$

option (c) is the right answer

Q9. Two poles of equal height are standing opposite to each other on either side of a road which is 100m wide. From a point between them on road , angle of elevation of their tops are and . The height of each pole (in meter) is

(a) 25 $\sqrt { 3 }$ (b) 20 $\sqrt { 3 }$

height and distance questions trigonometry application

In\quad \Delta AOC,\\ \\ tan60=\frac { AC }{ CO } =\frac { h }{ x } \\\ \\ \sqrt { 3 } =\frac { h }{ x } \\\ \\ x=\frac { h }{ \sqrt { 3 } } \quad …..(a)\\\ \\ \\\ \\ In\quad \Delta BOD,\\ \\ tan30=\frac { BD }{ BO } =\frac { h }{ 100-x } \\\ \\ =>\quad \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 100-x } \\\ \\ =>\quad 100-x=h\sqrt { 3 } \\\ \\ =>\quad 100-h\sqrt { 3 } =\quad x\\\ \\ =>\quad 100-h\sqrt { 3 } =\frac { h }{ \sqrt { 3 } } \quad \quad [from\quad 'a']\\\ \\ =>100=\frac { h }{ \sqrt { 3 } } +h\sqrt { 3 } \ \\\ \\ =>100=\frac { h+3h }{ \sqrt { 3 } } \\\ \\ =>100\times \sqrt { 3 } =4h\\\ \\ =>h=\frac { 100\times \sqrt { 3 } }{ 4 } \\\ \\ =>h=25\sqrt { 3 } \\\ \\ \\\ \\ \quad x=\frac { h }{ \sqrt { 3 } } =\frac { 25\sqrt { 3 } }{ \sqrt { 3 } } \\\ \\ x=25m\\ \\ OC=25m\quad and\quad OD=75m

Option (a) is the right answer

Height and Distance Quantitative Aptitude

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